擅长:python、mysql、java
<p>有了,itertools.groupby这个过程将更加简单:</p>
<pre><code>>>> key = operator.itemgetter(0)
>>> grouped = itertools.groupby(sorted(fileListCode, key=key), key=key)
>>> [(i, {k[1]: n for n, k in enumerate(j, 1)}) for i, j in grouped]
[('Seq3.xls', {'B08524_052': 1, 'B08524_053': 2, 'B08524_054': 3}),
('Seq98.xls', {'B25034_001': 1, 'B25034_002': 2, 'B25034_003': 3})]
</code></pre>
<p>对于旧版Python:</p>
^{pr2}$
<p>但我认为使用dict会更好:</p>
<pre><code>>>> {i: {k[1]: n for n, k in enumerate(j, 1)} for i, j in grouped}
{'Seq3.xls': {'B08524_052': 1, 'B08524_053': 2, 'B08524_054': 3},
'Seq98.xls': {'B25034_001': 1, 'B25034_002': 2, 'B25034_003': 3}}
</code></pre>